Find the value of : `sin{cos("cos"^(-1)(3pi)/4)}`

Explanation

To solve the problem \( \sin\left(\cos\left(\cos^{-1}\left(\frac{3\pi}{4}\right)\right)\right) \), we will follow these steps:

### Step 1: Understand the Inverse Cosine
Let \( x = \cos^{-1}\left(\frac{3\pi}{4}\right) \). By definition of the inverse cosine function, this means:
\[ \cos(x) = \frac{3\pi}{4} \]
However, we need to check if \( \frac{3\pi}{4} \) is a valid input for the cosine function. The range of the cosine function is between -1 and 1, and since \( \frac{3\pi}{4} \) is greater than 1, this indicates that we have made a mistake in interpreting the problem.

### Step 2: Re-evaluate the Input
Since \( \frac{3\pi}{4} \) is not a valid input for \( \cos^{-1} \), we need to reconsider the expression. The original question seems to be incorrectly stated. Assuming the intention was to find \( \sin\left(\cos\left(\cos^{-1}\left(\frac{3}{4}\right)\right)\right) \), we can proceed.

### Step 3: Define the New Variable
Let \( x = \cos^{-1}\left(\frac{3}{4}\right) \). Therefore:
\[ \cos(x) = \frac{3}{4} \]

### Step 4: Find \( \sin(x) \)
Using the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Substituting \( \cos(x) \):
\[ \sin^2(x) + \left(\frac{3}{4}\right)^2 = 1 \]
\[ \sin^2(x) + \frac{9}{16} = 1 \]
\[ \sin^2(x) = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \]
\[ \sin(x) = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 5: Find \( \sin\left(\cos(x)\right) \)
Now we need to find \( \sin\left(\cos(x)\right) \):
\[ \cos(x) = \frac{3}{4} \]
Thus, we need to find \( \sin\left(\frac{3}{4}\right) \).

### Step 6: Final Result
The final expression is:
\[ \sin\left(\cos\left(\cos^{-1}\left(\frac{3}{4}\right)\right)\right) = \sin\left(\frac{3}{4}\right) \]

### Conclusion
The value of \( \sin\left(\cos\left(\cos^{-1}\left(\frac{3}{4}\right)\right)\right) \) is \( \frac{\sqrt{7}}{4} \).